Trigonometry 

Reference 
Laws and theorems 
Calculus 
In trigonometry, the law of cotangents^{[1]} is a relationship among the lengths of the sides of a triangle and the cotangents of the halves of the three angles.
Just as three quantities whose equality is expressed by the law of sines are equal to the diameter of the circumscribed circle of the triangle (or to its reciprocal, depending on how the law is expressed), so also the law of cotangents relates the radius of the inscribed circle of a triangle (the inradius) to its sides and angles.
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Transcription
Determine the six trigonometric ratios for angle A in the right triangle below. So this right over here is angle A, it's at vertex A. And to help me remember the definitions of the trig ratios and these are human constructed definitions that have ended up being very, very useful for analyzing a whole series of things in the world. To help me remember them, I use the words soh cah toa. Let me write that down. Soh cah toa. Sometimes you can think of it as one word, but it's really the three parts that define at least three of the trig functions for you. And then we can get the other three by looking at the first three. So soh tells us that sine of an angle in this case it's sine of A so sine of A is equal to the opposite, that's the O, over the hypotenuse. Well in this context, what is the opposite side to angle A? Well, we go across the triangle, it opens up onto side BC. It has length 12. So that is the opposite side. So, this is going to be equal to 12. And what's the hypotenuse? Well, the hypotenuse is the longest side of the triangle. It's opposite the 90 degree angle. And so we go opposite the 90 degree angle, longest side is side AB. It has length 13. So this right over here is the hypotenuse. So, the sine of A is 12/13. Now let's go to cah. Cah defines cosine for us. It tells us that cosine of an angle in this case, cosine of A is equal to the adjacent side to the angle over the hypotenuse. So, what's the adjacent side to angle A? Well, if we look at angle A, there are two sides that are next to it. One of them is the hypotenuse. The other one has length 5. The adjacent one is side CA. So it's 5. And what is the hypotenuse? Well, we've already figure that out. The hypotenuse is right over here, it's opposite the 90 degree angle. It's the longest side of the right triangle. It has length 13. So the cosine of A is 5/13. And let me label this. This right over here is the adjacent side. And this is all specific to angle A. The hypotenuse would be the same regardless of what angle you pick, but the opposite and the adjacent is dependent on the angle that we choose in the right triangle. Now let's go to toa. Toa defines tangent for us. It tells us that the tangent of an angle is equal to the opposite side over the adjacent side. So given this definition, what is the tangent of A? Well, the opposite side, we already figured out, has length 12. And the adjacent side, we already figure out, has length 5. So the tangent of A, which is opposite over adjacent, is 12/5. Now, we'll go the to the other three trig ratios, which you could think of as the reciprocals of these right over here. But I'll define it. So first you have cosecant. And cosecant, it's always a little bit unintuitive why cosecant is the reciprocal of sine of A, even though it starts with a co like cosine. But, cosecant is the reciprocal of the sine of A. So sine of A is opposite over hypotenuse. Cosecant of A is hypotenuse over opposite. And so what's the hypotenuse over the opposite? Well, the hypotenuse is 13 and the opposite side is 12. And notice that 13/12 is the reciprocal of 12/13. Now, secant of A is the reciprocal. So instead of being adjacent over hypotenuse, which we got from the cah part of soh cah toa, it's hypotenuse over adjacent. So what is the secant of A? Well, the hypotenuse, we've figured out multiple times already, is 13. And what is the adjacent side? It's 5. So it's 13/5, which is, once again, the reciprocal of the cosine of A, 5/13. Finally, let's get the cotangent. And the cotangent is the reciprocal of tangent of A. Instead of being opposite over adjacent, it is adjacent over opposite. So what is the cotangent of A? Well, we've figured out the adjacent side multiple times for angle A. It's length 5. And the opposite side to angle A is 12. So it's 5/12, which is, once again, the reciprocal of the tangent of A, which is 12/5.
Contents
Statement
Using the usual notations for a triangle (see the figure at the upper right), where a, b, c are the lengths of the three sides, A, B, C are the vertices opposite those three respective sides, α, β, γ are the corresponding angles at those vertices, s is the semiperimeter, that is, s = a + b + c/2, and r is the radius of the inscribed circle, the law of cotangents states that
and furthermore that the inradius is given by
Proof
In the upper figure, the points of tangency of the incircle with the sides of the triangle break the perimeter into 6 segments, in 3 pairs. In each pair the segments are of equal length. For example, the 2 segments adjacent to vertex A are equal. If we pick one segment from each pair, their sum will be the semiperimeter s. An example of this is the segments shown in color in the figure. The two segments making up the red line add up to a, so the blue segment must be of length s − a. Obviously, the other five segments must also have lengths s − a, s − b, or s − c, as shown in the lower figure.
By inspection of the figure, using the definition of the cotangent function, we have
and similarly for the other two angles, proving the first assertion.
For the second one—the inradius formula—we start from the general addition formula:
Applying to cot(α/2 + β/2 + γ/2) = cot π/2 = 0, we obtain:
(This is also the triple cotangent identity)
Substituting the values obtained in the first part, we get:
Multiplying through by r^{3}/s gives the value of r^{2}, proving the second assertion.
Some proofs using the law of cotangents
A number of other results can be derived from the law of cotangents.
 Heron's formula. Note that the area of triangle ABC is also divided into 6 smaller triangles, also in 3 pairs, with the triangles in each pair having the same area. For example, the two triangles near vertex A, being right triangles of width s − a and height r, each have an area of 1/2r(s − a). So those two triangles together have an area of r(s − a), and the area S of the whole triangle is therefore
 This gives the result
 S = √s(s − a)(s − b)(s − c)
 as required.
 Mollweide's first formula. From the addition formula and the law of cotangents we have
 This gives the result
 as required.
 Mollweide's second formula. From the addition formula and the law of cotangents we have
 Here, an extra step is required to transform a product into a sum, according to the sum/product formula.
 This gives the result
 as required.
 The law of tangents can also be derived from this (Silvester 2001, p. 99).
See also
References
 ^ The Universal Encyclopaedia of Mathematics, Pan Reference Books, 1976, page 530. English version George Allen and Unwin, 1964. Translated from the German version Meyers Rechenduden, 1960.
 Silvester, John R. (2001). Geometry: Ancient and Modern. Oxford University Press. p. 313. ISBN 9780198508250.